Question

What are the important points needed to graph `y=3x^2+6x-1`?

Answer 1

Vertex: `(-1, -4)` , axis of symmetry: `x=-1`, x-intercepts:`x~~ -2.155 and x ~~ 0.155`, y-intercept:
`y=-1`, additional points:`(1,8)and (-3,8)`

Explanation:

This is equation of parabola , so vertex , axis of symmetry,

x intercepts , y intercept , opening of parabola , additional points

on the parabola are needed to draw graph.

`y=3 x^2+6 x-1 or y=3(x^2+2 x)-1` or

`y=3(x^2+2 x+1)-3-1 or 3(x+1)^2 -4`

This is vertex form of equation ,`y=a(x-h)^2+k ; (h,k)`

being vertex , here `h=-1 ,k=-4,a=3` Since `a` is positive,

parabola opens upward and vertex is at `(-1, -4)`.

Axis of symmetry is `x= h or x = -1 ;`

y-intercept is found by putting `x=0` in the equation

`y=3 x^2+6 x-1 :.y =-1 or (0,-1)`

x-intercepts are found by putting `y=0` in the equation

`0=3(x+1)^2 -4 or 3(x+1)^2 = 4` or

`(x+1)^2 = 4/3 or (x+1) = +- 2/sqrt3 or x = -1+- 2/sqrt 3`

or `x~~ -2.155 and x ~~ 0.155` . Additional points:

`x= =1 :. y=3(1+1)^2=8 or (1,8)` and

`x= =-3 :. y=3(-3+1)^2=8 or (-3,8)`

graph{3x^2+6x-1 [-10, 10, -5, 5]} [Ans]